How to delete the first word in each line of a file?

Question

consider a file names 'file.txt'. It contains the following.

dove is a bird
tiger is an animal
cricket is a game.

Expected output:

is a bird
is an animal
is a game.

Answer 1

To do it using cut

cut -f 2- -d ' ' file.txt > new_file.txt

"Give me the second and any other field beyond, using space as a delimiter, from the file.txt file and direct the output to new_file.txt"

Answer 2

using sed :

sed 's/[^ ]* //' list.txt

• will remove all till first space

Answer 3

with awk:

awk '{ $1=""; print substr($0,2) }' file.txt

Answer 4

With sed

To remove the first word: sed -E "s,^[[:alnum:]]+ ,," list.txt

To remove the first character: sed -E "s,^[[:alnum:]],," list.txt

Answer 5

In Perl:

perl -pe 's/.*? //'  input

In Grep:

grep -Po ' \K.*' input

Answer 6

Use ed, man!

ed -s input <<< $'%s/^[^ ][^ ]* //\nw\nq'

or, with a here-string:

printf '%s\n' '%s/^[^ ][^ ]* //' 'w' 'q' | ed -s input

This sends three newline-separated commands to ed:

1. on every (%) line, search and replace one or more non-space characters and a trailing space with nothing; the search pattern is anchored to match at the beginning of the line with ^
2. write the file back to disk
3. quit

Answer 7

awk is best, because it also removed preceding whitespace. And it's aliasable:

alias removefirstword="awk '{ \$1=\"\"; print substr(\$0,2) }'"

testhost: uptime

15:26:50 up 0 days, 14:42, 1 users, load average: 1.24, 1.28, 1.33

testhost: uptime | removefirstword

up 0 days, 14:42, 1 users, load average: 1.19, 1.26, 1.32

Of course, the other regexp solutions could do this, too.

Answer 8

With bash:

while read; do
  set -f -- $REPLY
  shift
  echo "$@"
done < infile

Output:

is a bird
is an animal
is a game.